# 14.2: C.2- 1st-order isotropic tensors

- Page ID
- 18092

A 1st-order tensor is a vector. If the vector \(\vec{v}\) is isotropic, then under an infinitesimal rotation

\[v_{i}^{\prime}=v_{j} C_{j i}=v_{j}\left(\delta_{j i}+r_{j i}\right)=v_{i}+v_{j} r_{j i}=v_{i}.\]

Therefore,

\[v_{j} r_{j i}=0.\]

This represents three algebraic equations, one for each value of \(i\):

\[\begin{array}{l}

v_{1} r_{11}+v_{2} r_{21}+v_{3} r_{31}=0 \\

v_{1} r_{12}+v_{2} r_{22}+v_{3} r_{32}=0 \\

v_{1} r_{13}+v_{2} r_{23}+v_{3} r_{33}=0.

\end{array}\label{eqn:1}\]

Because \(\underset{\sim}{r}\) is antisymmetric, \(r_{11}\) = \(r_{22}\) = \(r_{33}\) = 0, removing one term from each equation.

Now consider the first equation of Equation \(\ref{eqn:1}\):

\[v_{2} r_{21}+v_{3} r_{31}=0.\label{eqn:2}\]

Here is a crucial point: **if \(\vec{v}\) is isotropic, then Equation \(\ref{eqn:1}\) must be true for all antisymmetric matrices** \(\underset{\sim}{r}\), i.e., regardless of the values of \(r_{21}\) and \(r_{31}\). The only way this can be true is if \(v_2\) = 0 and \(v_3\) = 0. The same considerations applied to the second equation of Equation \(\ref{eqn:1}\) tell us that \(v_1\) must also be zero, hence the only isotropic 1st-order tensor is the trivial case

\[\vec{v}=0.\]